${3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}$
$+ {2 K}_{2} S O_{4} + 11 H_{2} O$

Equivalent mass of $K_{2} {C r}_{2} O_{7}$ for the above reaction is :

\frac{m o l a r m a s s}{6}

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\frac{m o l a r m a s s}{3}

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\frac{m o l a r m a s s}{2}

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\frac{m o l a r m a s s}{1}

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Solution

The correct option is A $\frac{m o l a r m a s s}{6}$
${3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3} + {2 K}_{2} S O_{4} + 11 H_{2} O$
+6 +3
Change in oxidation state of each Cr $= 3$ ,
As each $K_{2} {C r}_{2} O_{7}$ contains 2 moles of Cr. So, total change in oxidation is 6.
Therefore, equivalent weight of $K_{2} {C r}_{2} O_{7} = \frac{M}{6}$ .

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Similar questions

{3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}

+ {2 K}_{2} S O_{4} + 11 H_{2} O

{C H}_{3} {C H}_{2} O H

({C r}_{2} O_{7}^{- 2})

{C r}^{+ 3}

K_{2} {C r}_{2} O_{7}

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

1^{\circ}

1) E t - C O O H i . B H_{3} - ------ \to i i . H^{+} / H_{2} O

2) (C H_{3})_{2} C H - C O O H N a B H_{4} - --- \to

3) E t - C O O C H_{3} H^{+} / H_{2} O - ---- \to

4) (C H_{3})_{2} C = C H_{2} i . B H_{3}, T H F - -------- \to i i . H_{2} O_{2}, O H^{-}

5) E t - C O C l H_{2} - ------ \to P d - B a S O_{4}

6) (C H_{3})_{2} C H - O - E t H^{+} / H_{2} O - ---- \to

(C H_{3})_{2} C O + H C N \to X H_{3} O^{+} - -- \to Y H_{2} S O_{4} - --- \to H e a t Z

Z

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